For any three vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ prove that [$\vec{a}$ + $\vec{b}$ $\vec{b}$ + $\vec{c}$ $\vec{c}$ + $\vec{a}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{c}$]

Solution :

We have [$\vec{a}$ + $\vec{b}$ $\vec{b}$ + $\vec{c}$ $\vec{c}$ + $\vec{a}$]

= {($\vec{a}$ + $\vec{b}$)$\times$($\vec{b}$ + $\vec{c}$)}.($\vec{c}$ + $\vec{a}$)

= {$\vec{a}$$\times$$\vec{b}$ + $\vec{a}$$\times$$\vec{c}$ + $\vec{b}$$\times$$\vec{b}$ + $\vec{b}$$\times$$\vec{c}$}.($\vec{c}$ + $\vec{a}$)  {$\vec{b}$$\times$$\vec{b}$ = 0}

= {$\vec{a}$$\times$$\vec{b}$ + $\vec{a}$$\times$$\vec{c}$ + $\vec{b}$$\times$$\vec{c}$}.($\vec{c}$ + $\vec{a}$)

= ($\vec{a}\times\vec{b}$).$\vec{c}$ + ($\vec{a}\times\vec{c}$).$\vec{c}$ + ($\vec{b}\times\vec{c}$).$\vec{c}$ + ($\vec{a}\times\vec{b}$).$\vec{a}$ + ($\vec{a}\times\vec{c}$).$\vec{a}$ + ($\vec{b}\times\vec{c}$).$\vec{a}$

= [$\vec{a}$ $\vec{b}$ $\vec{c}$] + 0 + 0 + 0 + 0 + [$\vec{b}$ $\vec{c}$ $\vec{a}$]

= [$\vec{a}$ $\vec{b}$ $\vec{c}$] + [$\vec{a}$ $\vec{b}$ $\vec{c}$] = 2[$\vec{a}$ $\vec{b}$ $\vec{c}$]

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